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Let $\Gamma $ be the circumcircle of triangle $ABC$. Let $A_0$ be the center of the circle lying outside of $\triangle ABC$ and which is tangent to the segment $BC$ and to rays $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Points $B_0$ and $C_0$ are similarly defined. Let $A_1$ be the second intersection point of line $AA_0$ with $\Gamma$. Points $B_1$ and $C_1$ are similarly defined.

Prove that the area of $\triangle A_0B_0C_0$ is equal to twice the area of hexagon $AB_1CA_1BC_1$.

Working through this I found the alternate segment theorem very useful as I was trying to establish congruence between triangles that are in the hexagon but outside the triangle. I couldn't quite get there. Any help would be greatly appreciated.

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2 Answers 2

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Figure 1

Let $I$ be the point common to the three angle bisectors. (Note that $I$ is the center of the triangle's incircle (not shown); it is not the center of the circumcircle.)

Claim: $A_1$ is the midpoint of $\overline{IA_0}$.

Given the claim, we must have $|\triangle ICA_0| = 2 |\triangle ICA_1|$, because the "base" $\overline{IA_0}$ of the first triangle has twice the length of the "base" $\overline{IA_1}$ of the second, whereas their "heights" to $C$ match. Since $\triangle ICA_0$ and its counterparts throughout the figure cover $\triangle A_0B_0C_0$, while $\triangle ICA_1$ and its counterparts cover hexagon $AB_1CA_1BC_1$, the triangle's area must be twice the hexagon's.

Proof of claim:

Since the vertical angles $\angle AIC_1$ and $CIA_1$ subtend circumcircle arcs $\stackrel{\frown}{AC_1}$ and $\stackrel{\frown}{CA_1}$ that are respectively subtended by inscribed angles $\angle ACC_1$ and $\angle CAA_1$, their measure is equal to the sum of the measures of those inscribed angles:

Figure 2


Since $\angle A_1AB$ and $\angle A_1CB$ subtend arc $\stackrel{\frown}{A_1B}$, they are congruent:

Figure 3


Consequently, $\triangle A_1IC$ has congruent angles, $\angle I \cong \angle C$, and thus also congruent sides, $\overline{A_1I}\cong\overline{A_1C}$.

Figure 4

Note that, just as incenter $I$ lies on $\triangle ABC$'s three (interior) angle bisectors, excenter $A_0$ lies on two exterior angle bisectors; likewise for $B_0$ (and $C_0$), so that $\overline{A_0B_0}$ must contain $C$ and be perpendicular to $\overline{CC_0}$. We find, then, that $\angle IA_0C$ and $\angle A_1CA_0$ are complements of congruent angles and must be congruent to each other, whence $\overline{A_0A_1}\cong\overline{A_1C}$ ($\cong\overline{A_1I}$, from above). Point $A_1$ is therefore the midpoint of $\overline{IA_0}$ as claimed, completing the proof. $\square$

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Hint: The circumcenter and executer lies on the respective angle bisectors of the triangle. Let $O$ be the circumcenter of $ABC$, and $I$ be the incenter of $ABC$.

Hint: Consider right triangle $ICA_0$. Show that triangles $IA_1 C$ and $CA_1 A_0$ are isosceles, which gives that $IA_1 = A_1 C = A_1A_0$.

Hint: Decompose the hexagon into 3 parts, and the triangle into 3 parts, to show that the area is twice.

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How do we know that those lines are the angle bisectors of the triangle??? –  John Marty Apr 13 '13 at 11:14
    
@JohnMarty: I believe Calvin may have mis-typed in his first hint. The (centers of the ) in-circle and excircles lie on angle bisectors; note that the centers of excircles also lie on bisectors of external angles, so that, for instance $A_0$, $B$, and $C_0$ are collinear, and their line makes a right angle with the (internal) angle bisector at $B$. –  Blue Apr 13 '13 at 11:54
    
Thanks: I thought their was something wrong –  John Marty Apr 13 '13 at 11:59
    
@JohnMarty: Also, I believe that Calvin's second hint is meant to show $IA_1 = CA_1 = A_0A_1$. Specifically, if you can show that $A_1$ is the midpoint of $IA_0$, then you're done (because then $2|\triangle ICA_1| = |\triangle ICA_0|$, and likewise around the figure). –  Blue Apr 13 '13 at 12:04
    
@JohnMarty Sorry, edits made. –  Calvin Lin Apr 13 '13 at 16:15

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